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\begin{document}

\title{高等代数一}
\subtitle{1-克拉默公式、行列式、余子式、代数余子式}
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW} }
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{ {\ppr 2022年9月20日} }

\maketitle

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\begin{frame}{目录 }

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\begin{enumerate}
\item  线性方程组的求解
\item 克拉默公式
\item 行列式的按第一列展开
\item 余子式
\item 代数余子式
\end{enumerate}

\vfill 

主要内容：导出线性方程组的克拉默公式，按照第一列展开行列式，计算行列式的余子式与代数余子式。


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\begin{itemize}
\item 例子1：找到两个数 $A,B$，使得下述线性方程组的两个方程，分别乘以这两个数，相加之后可以消去未知数 $x_2$,
\begin{eqnarray*}
\left\{\begin{array}{rcl}
x_1+2x_2&=&3, \\
2x_1+3x_2&=&5.
\end{array}\right.
\end{eqnarray*}

\item 解答：可选 $A=3$ 与 $B=-2$, 
\begin{eqnarray*}
 x_1 +2x_2 &=& 3 \hspace{2cm} \times (3) \\
2x_1+3x_2&=& 5 \hspace{2cm}  \times (-2) \\
(3-4)x_1 + (6-6)x_2 &=& 9-10
%x_1 +0x_2 & = & 1 
\end{eqnarray*}

\item 求得 $$x_1=\frac{3A+5B}{A+2B}=\frac{9-10}{3-4}=1. $$
\end{itemize}

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\begin{frame}{1.2. 克拉默公式（二阶）}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}
\item 考虑下述线性方程组，
\begin{eqnarray*}
\left\{\begin{array}{rcl}
ax+by&=&{\color{red}m},  \hspace{2cm} \times (d) \\
cx+dy&=&{\color{red}n}, \hspace{2.1cm} \times (-b)
\end{array}\right.
\end{eqnarray*}

\item 使用二阶行列式的记号，则求解的克拉默公式为
\begin{eqnarray*}
x= \frac{md-nb}{ad-cb}=: \frac{\begin{vmatrix} {\color{red}m}&b \\ {\color{red}n}&d \end{vmatrix}}{\begin{vmatrix} a&b \\ c&d \end{vmatrix}}, \hspace{0.3cm}
y= \frac{md-nb}{ma-nc}=: \frac{\begin{vmatrix} a&{\color{red}m} \\ c&{\color{red}n} \end{vmatrix}}{\begin{vmatrix} a&b \\ c&d \end{vmatrix}}.
\end{eqnarray*}

\end{itemize}

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\begin{frame}{1.3. 线性方程组的求解（三阶）}

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\begin{itemize}
\item 例子2：找到三个数，使得下述线性方程组的三个方程，分别乘以这三个数，相加之后可以{\color{red}同时消去}未知数 $x_2,x_3$,
\begin{eqnarray}
%\left\{\begin{array}{rcl}
x_1+2x_2+3x_3&=&4, \hspace{2cm} \times (A) \\
5x_1+x_2+2x_3&=&3, \hspace{2cm} \times (B) \\
x_1+3x_2+x_3&=&2. \hspace{2cm} \times (C) 
%\end{array}\right.
\end{eqnarray}
\item 解答：一种方法是使用待定系数法，设所求的三个数为 $A,B,C$, 则为了同时消去未知数 $x_2,x_3$, 必须有%（当 $C\neq 0$）
\begin{eqnarray*}
\left\{\begin{array}{rcl}
2A+B+3C &=& 0 \\
3A+2B+C &=& 0
\end{array}\right.
\overset{\text{设 $C$不等于零}}{\Rightarrow}
\left\{\begin{array}{rcl}
2(\frac{A}{C})+(\frac{B}{C}) &=& -3 \\
3(\frac{A}{C})+2(\frac{B}{C})&=& -1
\end{array}\right.
\end{eqnarray*}

\item 这就归结为两个未知数两个方程的求解。
\end{itemize}

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\begin{itemize}
\item 另一种方法，根据例子1的求解，可以预见到{\color{red}所求的 $A$ 将与第 (2)(3) 个方程的 $x_2, x_3$ 的四个系数都有关}。
因此猜测
\begin{eqnarray*}
A=\begin{vmatrix} 1&2 \\ 3&1 \end{vmatrix}=-5, \hspace{0.5cm}
B=\begin{vmatrix} 2&3 \\ 3&1 \end{vmatrix}=-7, \hspace{0.5cm}
C=\begin{vmatrix} 2&3 \\ 1&2 \end{vmatrix}=1. 
\end{eqnarray*}

\item 测试这三个数是否能够同时消去两个未知数，
\begin{eqnarray}
\left\{\begin{array}{rcl}
x_1+2x_2+3x_3&=&4, \hspace{2cm} \times (-5) \\
5x_1+x_2+2x_3&=&3, \hspace{2cm} \times (-7) \\
x_1+3x_2+x_3&=&2. \hspace{2cm} \times (1) 
\end{array}\right.
\end{eqnarray}

\item 观察到 $B$ 取 $7$ 而不是 $-7$ 的时候，正好实现目标，此时，
$$x_1 = \frac{4A+3B+2C}{A+5B+C}=\frac{3}{29}. $$ 
\end{itemize}

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\begin{itemize}

\item {\color{red}上述计算过程蕴含了定义行列式与导出克拉默公式的基本思想。}

\item 考虑下述线性方程组，
\begin{eqnarray*}
\left\{\begin{array}{rcl}
a_1x_1+a_2x_2+a_3x_3&=&k,  \hspace{2cm} \times (A=\begin{vmatrix} b_2&b_3 \\ c_2&c_3 \end{vmatrix}) \\
b_1x_1+b_2x_2+b_3x_3&=&m,  \hspace{1.9cm} \times (B=-\begin{vmatrix} a_2&a_3 \\ c_2&c_3 \end{vmatrix}) \\
c_1x_1+c_2x_2+c_3x_3&=&n,  \hspace{2cm} \times (C=\begin{vmatrix} a_2&a_3 \\ b_2&b_3 \end{vmatrix}) \\
\end{array}\right.
\end{eqnarray*}

\item 适当选取三个数 $A,B,C$, 可得未知数 $x_1$ 的求解公式为
\begin{eqnarray*}
x_1 = \frac{kA+mB+nC}{a_1A+b_1B+c_1C}
\end{eqnarray*}

\end{itemize}


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\begin{frame}{1.6. 三阶行列式的诞生}

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\begin{itemize}
\item 将上述 $x_1$ 的的求解公式的分母
\begin{eqnarray}
{\color{red}a_1}A + {\color{red}b_1}B + {\color{red}c_1}C=
 {\color{red}a_1} \begin{vmatrix} b_2&b_3 \\ c_2&c_3 \end{vmatrix} 
+ {\color{red}b_1} (-\begin{vmatrix} a_2&a_3 \\ c_2&c_3 \end{vmatrix}) 
+ {\color{red}c_1} \begin{vmatrix} a_2&a_3 \\ b_2&b_3 \end{vmatrix}
\label{key-expression}
\end{eqnarray}
写成一个三阶行列式的形式
\begin{eqnarray}
\begin{vmatrix} {\color{red}a_1}&a_2&a_3 \\ {\color{red}b_1}&b_2&b_3 \\ {\color{red}c_1}&c_2&c_3 \end{vmatrix}.
\label{key-determinant}
\end{eqnarray}

\item {\color{red}将表达式 (\ref{key-expression}) 称为这个三阶行列式 (\ref{key-determinant}) 的按第一列展开。}

\item 未知数 $x_1$ 是两个三阶行列式的商。而且这两个三阶行列式非常像。
\end{itemize}

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\begin{frame}{1.7. 克拉默公式（三阶）}

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\begin{itemize}

\item 考虑下述线性方程组，
\begin{eqnarray*}
\left\{\begin{array}{rcl}
{\color{red}a_1}x_1+a_2x_2+a_3x_3&=&{\color{blue}k},\\
{\color{red}b_1}x_1+b_2x_2+b_3x_3&=&{\color{blue}m}, \\
{\color{red}c_1}x_1+c_2x_2+c_3x_3&=&{\color{blue}n}. 
\end{array}\right.
\end{eqnarray*}

\item 未知数 $x_1$ 的求解公式如下，未知数 $x_2$ 与 $x_3$ 的公式类似。
\begin{eqnarray*}
x_1 = \frac{{\color{blue}k}A+{\color{blue}m}B+{\color{blue}n}C}
{{\color{red}a_1}A + {\color{red}b_1}B + {\color{red}c_1}C} = 
\frac{\begin{vmatrix} {\color{blue}k}&a_2&a_3 \\ {\color{blue}m}&b_2&b_3 \\ {\color{blue}n}&c_2&c_3 \end{vmatrix}}
{\begin{vmatrix} {\color{red}a_1}&a_2&a_3 \\ {\color{red}b_1}&b_2&b_3 \\ {\color{red}c_1}&c_2&c_3 \end{vmatrix}}.
\end{eqnarray*}

\end{itemize}

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\begin{frame}{1.8. 四阶行列式的定义}

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\begin{itemize}

\item {\color{red}定义：四阶行列式按照第一列展开的结果，定义为由四个三阶行列式的组合得到的一个数学表达式，}
\begin{eqnarray*}
\begin{vmatrix} {\color{red}a_1}&a_2&a_3&a_4 \\ {\color{red}b_1}&b_2&b_3&b_4 \\ {\color{red}c_1}&c_2&c_3&c_4 \\ {\color{red}d_1}&d_2&d_3&d_4  \end{vmatrix}
={\color{red}a_1}A + {\color{red}b_1}B + {\color{red}c_1}C + {\color{red}d_1}D,
\end{eqnarray*}
其中 $A,B,C,D$ 是这个四阶行列式划去第1列，以及分别划去第1、2、3、4行之后得到的四个三阶行列式，再加上$+,-,+,-$ 这四个正负符号得到的表达式。


\end{itemize}

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\begin{itemize}

\item 下述线性方程组
\begin{eqnarray*}
\left\{\begin{array}{rcl}
{\color{red}a_1}x_1+a_2x_2+a_3x_3+a_4x_4&=&{\color{blue}k},  \hspace{2cm} \times (A) \\
{\color{red}b_1}x_1+b_2x_2+b_3x_3+b_4x_4&=&{\color{blue}m},  \hspace{1.9cm} \times (B) \\
{\color{red}c_1}x_1+c_2x_2+c_3x_3+c_4x_4&=&{\color{blue}n},  \hspace{2cm} \times (C) \\
{\color{red}d_1}x_1+d_2x_2+d_3x_3+d_4x_4&=&{\color{blue}p},  \hspace{2cm} \times (D) 
\end{array}\right.
\end{eqnarray*}
的克拉默公式为
\begin{eqnarray*}
x_1= \frac{{\color{blue}k}A + {\color{blue}m}B + {\color{blue}n}C + {\color{blue}p}D}{{\color{red}a_1}A + {\color{red}b_1}B + {\color{red}c_1}C + {\color{red}d_1}D} 
= \frac{{\text{\small\lib 将系数行列式的第一列换成常数列}}} {{\text{\small\lib 系数行列式}}}. 
\end{eqnarray*}

\end{itemize}

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\begin{frame}{1.10. 一些注释与疑问}

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\begin{itemize}

\item  二阶行列式展开得到2项，三阶行列式展开得到6项，四阶行列式展开得到24项。一般的 $n$ 阶行列式展开得到 $n!=n\cdot (n-1)\cdots 3\cdot 2\cdot 1$ 项。
\item  如何确定每一项前面的正负符号？
\item  如果一个线性方程组的系数行列式的值等于零，又该如何求解？
\item  如果一个线性方程组的未知数个数与方程个数不一样，又该如何求解？

\end{itemize}

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\begin{frame}{1.11. 余子式、代数余子式}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item {\color{red}定义：在 $n$ 阶行列式 
$$
D = \begin{vmatrix} 
a_{11} & a_{12} & \cdots & a_{1n} \\    
a_{21} & a_{22} & \cdots & a_{2n} \\    
\cdots & \cdots & \cdots & \cdots  \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\    
\end{vmatrix} =: |a_{ij}|_{n\times n} 
$$ 
中，划去元素 $a_{ij}$ 所在的行与列，剩下的 $n-1$ 阶行列式称为这个元素所对应的余子式，记为 $M_{ij}$.}
\item {\color{red}定义：余子式加上适当的正负号，称为代数余子式，具体来讲就是}
$${\color{red}A_{ij}=(-1)^{i+j}M_{ij}.}$$


\end{itemize}

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\begin{itemize}

\item 例子3：已知行列式 $D=\begin{vmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{vmatrix}$，求 $M_{23}$ 与 $A_{23}$. 
%\begin{eqnarray*}
%D=\begin{vmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{vmatrix}.
%\end{eqnarray*}

\item 解答：划去行列式 $D$ 的第2行与第3列，可得余子式 $$M_{23}=\begin{vmatrix} 1&2 \\  7&8 \end{vmatrix}=-6.$$ 
由此得到代数余子式 $$A_{23}=(-1)^{2+3}M_{23}=6. $$ 


\end{itemize}

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\begin{enumerate}
\item  使用克拉默公式，求线性方程组的第一个未知数 $x_1$ 的值，
\begin{eqnarray*}
\left\{\begin{array}{rcl}
x_1 + x_2 + 2x_3 + 3x_4&=&1, \\
3x_1 - x_2 - x_3 -2x_4&=&-4, \\
2x_1 + 3x_2 - x_3 - x_4&=&-6, \\
x_1 + 2x_2 + 3x_3 - x_4&=&-4. 
\end{array}\right.
\end{eqnarray*}

\item  已知行列式 $D=\begin{vmatrix} 1&2&3&4 \\ 5&6&7&8 \\ 9&0&1&2 \\ 3&4&3&4 \end{vmatrix}$，求余子式 $M_{34}$ 与代数余子式 $A_{34}$. 

\end{enumerate}

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\begin{enumerate}

\item  $x_1=-1$. 

\item  $M_{34}=8$, $A_{34}=-8$.

\end{enumerate}

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